Revealed reputations in the finitely repeated prisoners’ dilemma

Abstract

In a sequential-move, finitely repeated prisoners’ dilemma game (FRPD), cooperation can be sustained if the first-mover believes her opponent might be a behavioral type who plays a tit-for-tat strategy in every period. We test this theory by revealing second-mover histories from an earlier FRPD experiment to their current opponent. Despite eliminating the possibility of reputation-building, aggregate cooperation actually increases when histories are revealed. Cooperative histories lead to increased trust, but negative histories do not cause decreased trust. We develop a behavioral model to explain these findings.

JEL Classification

Notes

Acknowledgments

The authors thank Yaron Azrieli, Lucas Coffman, Glenn Dutcher, John Kagel, Semin Kim, Peter McGee, Xiangyu Qu, Arno Riedl, Dan Schley, Mike Sinkey, Tom Wilkening, and Chao Yang for their helpful comments and conversations. We are also grateful to two anonymous referees whose comments and insights have helped us to significantly improve the paper. Healy acknowledges financial support from National Science Foundation Grant #SES-0847406. Any opinions, findings, conclusions, or recommendations expressed are those of the authors and do not necessarily reflect the views of the Federal Trade Commission.

Proposition 4

Let \(p\in (0,1)\) be the common belief that the other player plays tit for tat and \(\overline{p}_t\) the period \(t\) posterior belief. The following is a sequential equilibrium for a sequential-move FRPD.

Proof

It is easier notationally to derive the equilibrium by counting backwards with \(t=10\) representing the first round of the supreme. In the body of the paper, however, time is indexed forward with \(t=1\) representing the first round of the supreme. Now, in any period \(t\), the first-mover will cooperate if

where \(V_{t-1}\left( p\right) \) is the continuation value of the first-mover entering period \(t-1\) with belief \(p\). Let \(V_{0}\equiv 0\). Let \(\bar{p}_{t}\) be the smallest value of \(p_{t}\) satisfying this inequality. (We will show later that the inequality in fact grows in \(p_{t}\).)

Table 9

Effect of player types on first-round cooperation and rounds of cooperation before first defection—types defined by last block 1 Supergame

1st round cooperation by first-mover

No. of rounds of cooperation before 1st defection

(i)

(ii)

(iii)

(iv)

(v)

Avg. second-mover Block 1 coop.

0.042

(0.251)

\(\times \) S2

0.116

(0.013)

Defector

0.648

(0.046)

Imitator

1.388

1.264

2.698

0.996

(0.006)

(0.377)

(\({<}0.001\))

(0.588)

Cooperator

1.996

2.278

1.950

2.425

(\({<}0.001\))

(0.006)

(0.038)

(0.016)

Trusting

\(\times \) Defector

0.482

8.374

(0.282)

(\({<}0.001\))

\(\times \) Imitator

1.921

9.921

(\({<}0.001\))

(\({<}0.001\))

\(\times \) Cooperator

0.973

8.144

(0.065)

(\({<}0.001\))

\(\times \) Unobserved

1.961

(\({<}0.001\))

S2

\(\times \) Defector

4.701

\(-\)4.950

5.729

(\({<}0.001\))

(\({<}0.001\))

(\({<}0.001\))

\(\times \) Imitator

4.434

0.040

6.079

(0.061)

(0.939)

(0.042)

\(\times \) Cooperator

5.028

0.953

6.521

(0.004)

(0.556)

(0.004)

2S\(\times \)Trusting

\(\times \) Defector

\(-\)10.205

(\({<}0.001\))

\(\times \) Imitator

\(-\)4.520

(0.039)

\(\times \) Cooperator

\(-\)4.042

(0.003)

Supergame#

0.385

0.503

0.839

\(-\)0.402

4.517

(0.035)

(0.047)

(0.143)

(0.466)

(0.019)

(Supergame#)\(^2\)

\(-\)0.047

\(-\)0.062

\(-\)0.099

0.066

\(-\)0.596

(0.121)

(0.138)

(0.262)

(0.469)

(0.049)

Constant

\(-\)0.307

\(-\)1.617

\(-\)4.219

5.983

\(-\)11.873

(0.226)

(\({<}0.001\))

(\({<}0.001\))

(\(<\)0.001)

(\(<0.001\))

Observations

660

660

660

405

255

Avg. second-mover Block 1 coop. represents the average number of rounds of cooperation in all of the Block 1 supergames for the second-mover. Column (iv) includes observations with Trusting first-mover and column (v) includes observations with Non-Trusting first-mover only. \(p\) values from robust standard errors clustered by individual first-mover

The probability a selfish second-mover cooperates is the highest \(q\) such that the first-mover is willing to cooperate in periods \(t-1\) after observing cooperation in period \(t\). Thus, if \(\bar{p}_{t}\) is the lowest belief at which first-mover will cooperate in period \(t\), then \(q_{t}\left( p\right) \) solves

For completeness, let \(q_{t}\left( 1\right) =1\) for all \(t\) and \(q_{t}\equiv 1 \) for any \(t\) where \(\bar{p}_{t-1}=0\). Since a selfish second-mover never cooperates in the last period, set \(q_{1}\left( p\right) =0\) for all \(p\). (This is equivalent to setting \(\bar{p}_{0}=1\).)

For any \(t>1\), consider the case where \(p_{t}\ge \bar{p}_{t-1}\). Here, \( q_{t}\left( p_{t}\right) =1\) (the second-mover cooperates with certainty) and

First-movers will cooperate in period \(T\) if \(p_{T}^{*}\ge \bar{p}_{T}\). Thereafter, they will cooperate as long as they have never seen a defection and will never cooperate after seeing a defection. In that case, beliefs will evolve according to the formula

Beliefs change to \(p_{t}^{*}=0\) if a defection is observed in any previous period. If \(p_{T}^{*}<\bar{p}_{T}\), then both players always defect and \(p_{t}^{*}=p_{T}^{*}\) for every period \(t\). The on-path continuation value of the first-mover will equal

Proof of Proposition 1

(a) Let the first-mover’s expected payoff in round \(s\) from the remaining rounds \(1,\ldots ,s\) given beliefs \(p_{1},\ldots ,p_{s}\) be denoted by \(V_{s}(p_{1},\ldots ,p_{s})\). The expected payoff for cooperating in round \(t\) is \(p_{t} (7 + V_{t-1}(p_{1},\ldots ,p_{t-1})) + (1 - p_{t}) V_{t-1}(0,\ldots ,0)\). The expected payoff for defecting in round \(t\), given that the second-mover will respond by defecting for at least one round, is at most \(4 + V_{t-1}(p_{1},\ldots ,p_{t} p_{t-1})\). Therefore, the first-mover plays tit for tat in period \(t\) if and only if the following inequality holds

The proof is by induction. First, we know that the first-mover cooperates in round \(1\) if and only if \(p_{1} 7 + (1 - p_{1}) 0 \ge 4\) holds. Therefore, if \(p_{1} \ge \frac{4}{7}\) holds, then we have \(V_{1}(p_{1}) = 7 p_{1}\), and if \(p_{1} < \frac{4}{7}\) holds, then we have \(V_{1}(p_{1}) = 4\), and the formula is true for \(t = 1\).

Now, assume that the formula holds for all rounds up to \(t-1\) and show that it holds for round \(t\). Assume that the following holds

(b) Let the second-mover’s expected payoff in round \(s\) from the remaining rounds \(1,\ldots ,s-1\) given beliefs \(p_{1},\ldots ,p_{s-1}\) be denoted by \(V_{s}(p_{1},\ldots ,p_{s-1})\). The expected payoff for cooperating in round \(t\) is \(7 + p_{t} (7 + V_{t}(p_{1},\ldots ,p_{t-1})) + (1 - p_{t}) (4 + V_{t}(0,\ldots ,0))\). The expected payoff for defecting in round \(t\), given that the first-mover will respond by defecting for at least one round, is at most \(12 + V_{t}(p_{1},\ldots ,p_{t} p_{t-1})\). Therefore, the second-mover plays tit for tat in period \(t+1\) if and only if the following inequality holds

The proof is by induction. First, we know that defection is the dominant action for the second-mover in round \(1\). The second-mover cooperates in round \(2\) if and only if \(7 + p_{1} 12 + (1 - p_{1}) 4 \ge 12 + 4\) holds. Therefore, if \(p_{1} \ge \frac{5}{8}\) holds, then we have \(V_{2}(p_{1}) = 4 + 8 p_{1}\), and if \(p_{1} < \frac{5}{8}\) holds, then we have \(V_{2}(p_{1}) = 4\), and the formula is true for \(t = 1\).

Now, we assume that the formula holds for all rounds up to \(t-1\) and show that it holds for round \(t\). Assume that the following holds

Proof of Proposition 2

By Proposition 1, the first-mover’s Block 1 strategy is \(s_{m}\) if and only if \(\mu \) is such that \(p_{k} \ge \frac{4}{\sum _{i=m+1}^{k} (3 \prod _{j=i}^{k-1} p_{j}) + 7 \prod _{i=m}^{k-1} p_{i}}\) holds for all \(k \in \{m,\ldots ,10\}\). This condition can be rewritten in terms of the prior beliefs \(\mu \) as

for all \(k \in \{m,\ldots ,10\}\). After several steps of algebra, the denominator of the right-hand side of the above inequality simplifies to \(\frac{1}{\sum _{i=1}^{k}\mu (s_{i})}((7 + 3(k - n))(\sum _{i=1}^{m}\mu (s_{i})) + 3\sum _{i=m+1}^{k}((k+1-i)\mu (s_{i})))\), and the condition can be simplified to

Now, suppose that the first-mover’s prior beliefs satisfy \(\mu (s_{k+1}) \le (3\big /4) \sum _{i=1}^{k} \mu (s_{i})\) for all \(k \in \{m,\ldots ,10\}\). For \(k=m\), the above condition is satisfied trivially. We now show that the above condition is satisfied for \(k=m+r\) for any \(r \ge 1\). For any \(r \ge 1\), the inequality \(\mu (s_{m+r+1}) \le (3\big /4) \sum _{i=1}^{m+r} \mu (s_{i})\) can be rewritten as

\(r \ge 2\) implies that \((\frac{3}{4})^{r-2} - r < 0\), so \(\delta < 0\) holds and the condition for the first-mover to play strategy \(s_{m}\) in Block 1 is satisfied.

Given that the second-mover always defected before her opponent by round \(n\) of Block 1 supergames, where \(m < n\), we have \(\tilde{\mu }(s_{k}) = 0\) for all \(k \le m\). Therefore, \(\tilde{p}_{l} = 0\) holds for all \(l \le m\). By Proposition 1, it follows that the first-mover’s Block 2 strategy is \(s_{m+t}\) for some \(t \ge 1\).

Proof of Proposition 3

(a) By Proposition 1, the first-mover’s Block 1 strategy is \(s_{n}\) if and only if \(\mu \) is such that \(p_{k} \ge \frac{4}{\sum _{i=n+1}^{k} (3 \prod _{j=i}^{k-1} p_{j}) + 7 \prod _{i=n}^{k-1} p_{i}}\) holds for all \(k \in \{n,\ldots ,10\}\). By similar logic to the proof of Proposition 2, if the first-mover’s prior beliefs satisfy \(\mu (s_{k+1}) \le (3\big /4) \sum _{i=1}^{k} \mu (s_{i})\) for all \(k \in \{n,\ldots ,10\}\), then the condition for the first-mover to play strategy \(s_{n}\) in Block 1 is satisfied. Given that the second-mover never defected before her opponent in rounds \(10,\ldots ,m\) of Block 1 supergames, where \(m < n\), we have \(\tilde{\mu }(s_{k}) = 0\) for all \(k > m\). Therefore, \(\tilde{p}_{l} = 1\) holds for all \(l \ge m\). By Proposition 1, it follows that the first-mover’s Block 2 strategy is \(s_{n-t}\) for some \(t \ge 1\).

(b) \(\mu (s_{11}) > 3\big /7\) implies that \(\mu (s_{11}) > (3\big /4) \sum _{i=1}^{10} \mu (s_{i})\) holds, so the condition for the first-mover to play strategy \(s_{11}\) in Block 1 is satisfied. Given that the second-mover never defected before her opponent in rounds \(10,\ldots ,m\) of Block 1 supergames, where \(m \le 10\), we have \(\tilde{\mu }(s_{k}) = 0\) for all \(k > m\). Therefore, \(\tilde{p}_{l} = 1\) holds for all \(l \ge m\). By Proposition 1, it follows that the first-mover’s Block 2 strategy is \(s_{11-t}\) for some \(t \ge 1\).